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I’m wondering if I’m making the right approach when I’m trying to calculate the power consumed by the diode in the circuit.

Here is the problem:

enter image description here

So my approach is:

Power loss by diode $$V_f \times I_f$$ where

\$V_f\$ = Diode forward voltage drop.

\$I_f\$ = The current flowing through the diode.


So the first thing I did was to calculate the current

$$I = V/R = 5/1000 = 5 mA$$

And then I use $$ V_f \times I_f= 2 \times 0.005 = 10 mW $$

Is this correct? I feel like there is more to the problem with the fact that it states ”red color light”. Should I check on a sheet or similar?

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    \$\begingroup\$ There are two problems with your analysis. The current is wrong (think about what V in that equation represents) and secondly 2 x 0.005 is not 10. \$\endgroup\$ – Spehro Pefhany Apr 5 '20 at 22:30
  • \$\begingroup\$ My hands went a little bit crazy while writing on my phone. Thank you for noting that. \$\endgroup\$ – Vetenskap Apr 5 '20 at 23:39
  • \$\begingroup\$ "red" has nothing to do with it. Or rather, yes red LEDs may be different from green or yellow or whatever, but it is thrown in there just to confuse you. \$\endgroup\$ – manassehkatz-Moving 2 Codidact Apr 6 '20 at 20:35
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I think that you have neglected the voltage drop across the diode which limits the current.

My solution is : $$ \because V_{R_1}=V_0-V_f=5-2=3\,\text{V} $$ $$ \therefore I_f=I_{R_1}=V_{R_1}/R_1=3/1000=3\,\text{mA} $$ $$Power Loss =V_f\times I_f=2 \times0.003=6\,\text{mW} $$

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You calculated the current wrong. Apply KVL. You will see that the voltage accross the resistor is actually 3 volts. Other than that, your approach seems correct.

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    \$\begingroup\$ Forget this edit, I'll put in my own answer instead. Sorry for wasting your time. \$\endgroup\$ – TpKnet Apr 6 '20 at 19:51
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You just forgot the voltage drop across the resistor so...

Power loss by diode $$V_f×I_f$$ where

\$V_f\$ = Diode forward voltage drop.

\$I_f\$ = The current flowing through the diode = current through the resistor.

So the first thing I did was to calculate the current through the resistor.

\$V_R=V_0-V_f = 5\,\text{V} - 2\,\text{V} = 3\,\text{V}\$.

\$I=V_R/R=3/1000=3\,\text{mA}\$

And then I use $$V_f×I_f=2×0.003=6\,\text{mW}$$

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\$V_R=V_0-V_f = 5 - 2 = 3\,\text{V}\$.

\$I=V_R/R=3/1000=3\,\text{mA}\$.

\$V_f×I_f=2×0.003=6\,\text{mW}\$.

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    \$\begingroup\$ Welcome to the site. Your answer needs to contain some explanations rather than just firing off the solution exam-style. It should be written clearly enough to teach future readers. Please can you edit it and greatly improve it. Thanks. \$\endgroup\$ – TonyM Apr 7 '20 at 10:22
  • \$\begingroup\$ Welcome to the site. I think your answer is correct but you'd better avoid answering a question which has several people answered the same answer. Thank you :). \$\endgroup\$ – NAND Apr 8 '20 at 17:52
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\$P_{diode} = 2\,\text{V} \times 0.003\,\text{A} = 6\,\text{mW}\$.

\$P_{resistor} = 3\,\text{V} \times 0.003\,\text{A} = 9\,\text{mW}\$.

\$P_{total} = 6\,\text{mW} + 9\,\text{mW} = 15\,\text{mW}\$ consumed for supplying the diode.

The efficiency of the system is $$ \frac{P_{diode}}{(P_{diode} + P_{resistor})} = \frac{6\,\text{mW}}{(6+9)\,\text{mW}} = 0.4 = 40\%. $$ Efficiency can be improved by using a supply with voltage closer to the diode voltage drop.
This will reduce the \$9\,\text{mW}\$ loss from heating the resistor.

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\$I_F = (V_O - V_F) / R_1 = 3 / 1k = 3 mA\$.

\$P_F = V_F \times I_F = 2 × 3 = 6 mW\$.

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    \$\begingroup\$ Welcome to electronics.SX! Your answer is correct, but when you see, that several people have already posted that same answer to the question before, it is not necessary to repost that again. Try to concentrate your efforts on questions without answers or questions where you think you can add new information to the existing answers. This will help other people and you will get upvotes for your answers! \$\endgroup\$ – jusaca Apr 7 '20 at 6:50
  • \$\begingroup\$ In second line, Did you mean \$R_F\$ or \$P_F\$? \$\endgroup\$ – NAND Apr 8 '20 at 17:28
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\$V = 5 - 0.7 - 2= 2.3\,\text{V}\$.

\$I = ?\$.

\$R = 1 \times 10^3\Omega\$.

\$P = V^2 ÷ R\$.

Hence:

\$ P = 2.3^2 ÷ 1000 = 5.29 × 10^-3\,\text{W}\approx 5\,\text{mW}\$.

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    \$\begingroup\$ There is no 0.7 V drop in the circuit. \$\endgroup\$ – Transistor Apr 7 '20 at 6:15
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First of all, led has voltage drop about 1.3V for red, 1.7 for green, 3.3 for White led etc. So on your dwg it is red so the current I=5-1,3=3.7V/1000ohms=3.7mA and so P=Vd*I=1.3*0.0037=2.9mW

Usually i set signali g led current to 10 mA, so in this case R=3.7/.01=370ohm. In this case I use nearest lower standard resistance i. e. 330ohm, so I=11mA and power 15mW.

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